Question

A diseased man marries a normal woman. They have three daughters and five sons. All the daughters were diseased and the sons were normal. The gene of this disease is -

• A. sex-linked recessive
• B. sex-linked dominant
• C. autosomal dominant
• D. sex-limited character

Answer 1

The correct option is B

sex-linked dominant

The correct option is B.

Explanation of correct option:

1. Daughters receive one X chromosome from each parent.
2. Here, all the daughters are diseased. Hence, the condition is an X-linked disease.
3. The father is diseased and therefore he will transmit the affected allele to the daughters. However, the mother is normal and doesn’t carry an allele for the disease.
4. Since the gene of the disease is expressed in the heterozygous condition, it is a sex-linked dominant gene.

Explanation of incorrect option:

Option A:

1. X-linked (sex-linked) recessive disorders mostly affect males since males are homozygous for the trait. Example: Colorblindness.

Option C:

1. In autosomal dominant disorders, the presence of one copy of the affected allele can cause the disease. It occurs due to mutation in autosomes. Example: Achondroplasia.

Option D:

1. Sex-limited genes are present in both sexes but it is expressed in members of only one sex. For example, premature baldness affects males.

Final answer:

A diseased man married to a normal woman has all diseased daughters but normal sons. The gene of this disease is sex-linked dominant.